In words, the sum of the product of the lengths of the opposite sides of a quadrilateral is at least the product of the lengths of its diagonals.

The equality for a cyclic quadrilateral is known as Ptolemy’s theorem while the more general inequality applying to any quadrilateral is called Ptolemy’s inequality. Many of the proofs below that establish Ptolemy’s theorem can be modified slightly to prove the inequality.

Let the diagonals intersect at and construct on the circumcircle so that is parallel to . A short angle chase shows that is an isosceles trapezium.

Then (angles on common arc and using ). Also triangles and have the same base and height and hence the same area.

Hence the area of ABCD is the sum of the areas of triangles ABE and ADE, or

where the last step follows from being an isosceles trapezium.

But also this area is and recall from above that . Therefore equating this with (1) we end up with .

Proof 4 (ref: [1])

Here three of the triangles of the cyclic quadrilateral are scaled to fit together to form a parallelogram. Namely,

is scaled by ,

is scaled by ,

is scaled by .

A simple angle chase based on the coloured angles shown reveals that a parallelogram is formed when the triangles are joined together. Ptolemy’s theorem then follows from equating two of its opposite side lengths.

Let the angles subtended by respectively be and let the circumradius of the quadrilateral be . By the sine rule , , , . Then the equality to be proved is equivalent to

But we have

as required.

Proofs of Ptolemy’s Inequality (all make use of the triangle inequality)

Proof 7:

Denote the points by vectors or complex numbers . Then we have the equality

Applying the modulus (length) to both sides and then the triangle inequality leads to

which is Ptolemy’s inequality.

Proof 8: (ref: [2])

Place the origin at the point so that are represented by the vectors respectively. Let . Then

Similarly, and . By the triangle inequality, , or in other words,

which is another way of writing Ptolemy’s inequality.

Proof 9: (inversion)

Recall that under inversion under a point a circle passing through maps to a line. If are points on the circle mapping to respectively under inversion in a circle of radius centred at , then

To see this, since by the definition of the inverse, . This together with the fact that angle is common implies that triangles and are similar. This gives us the relationship as desired.

For our quadrilateral apply an inversion centred at with radius . Then map to points which are collinear if is cyclic. Using the result above and similarly and . By the triangle inequality and so this becomes

which is Ptolemy’s inequality.

Proof 10: (ref: [3])

Construct on so that , then draw on with . Finally rotate about to map to .